Question

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Answer 1

Ответ:

1)

$a = log_{2}( log_{ \sqrt{2} }16 )$

$a = log_{2}( log_{ {2}^{ \frac{1}{2} } } {2}^{4} )$

$\frac{4}{ \frac{1}{2} } log_{2}(2)$

$\frac{4}{ \frac{1}{2} } \times 1$

$\frac{4}{ \frac{1}{2} } = 8$

$a = log_{2}(8)$

$a = log_{2}( {2}^{3} )$

$3 log_{2}(2)$

$3 \times 1$

$3$

$a = 3$

2)

$log_{2 \times 3}(x - 4) + log_{2 \times 3}(x + 1) = 1$

$x - 4 \leqslant 0 \\ x + 1 \leqslant 0$

$x \leqslant 4 \\ x \leqslant - 1$

$x\in < - \infty .4]$

$log_{2 \times 3}(x - 4) + log_{2 \times 3}(x + 1) = 1.x \in < 4. + \infty >$

$log_{6}(x - 4) + log_{6}(x + 1) = 1$

$log_{6}((x - 4)(x + 1)) = 1$

$log_{6}(x x + x - 4x - 4) = 1$

$log_{6}( {x}^{2} + x - 4x - 4 ) = 1$

${x}^{2} + x - 4x - 4 = {6}^{1}$

${x}^{2} + 1x - 4x - 4 = {6}^{1}$

${x}^{2} - 3x - 4 = 6$

${x}^{2} - 3x - 4 - 6 = 0$

${x}^{2} + 2x - 5x - 10 = 0$

$x(x + 2) - 5(x + 2) = 0$

$(x + 2)(x - 5) = 0$

$x + 2 = 0 \\ x - 5 = 0$

$x + 2 - 2 = 0 - 2 \\ x - 5 + 5 = 0 + 5$

$x = - 2 \\ x \in < 4. + \infty > \\ x = 5$

$x = 5$